Integration Question (1 Viewer)

[joesmith1975]

Find the area enclosed between the curve y=x(x^2 -2) and the x axis.


The answer is 2 units^2

Thanks in advance.

 

[Riviet]

y=x(x+sqrt2)(x-sqrt2)

So the x-intercepts are at x=0, sqrt2, and -sqrt2

Since the function is odd, you can take the integral from -sqrt2 to 0 of x³-2x (which should equal 1 unit²) and multiply it by two because A1=A2.

Hope the diagram helps.

 

[Sober]

Find the area enclosed between the curve y=x(x^2 -2) and the x axis.


The answer is 2 units^2

Thanks in advance.

Find the x-intercepts: x = 0, √2, -√2;

So we have to solve two separate areas and ensure they are positive by using absolute values, but if you notice that it is an odd function then you can only find the one on the right of the y axis and double it to find the other, here is the equation:

√2
/
| (x^3 - 2x) dx * 2 = 2 * [x^4 / 4 - x^2]~ = (1 - 2)*2 = -2
/
0

We want the positive area so absolute it, 2 uinits squared.

EDIT: damn time and its consistant cronology Riviet!

 

[joesmith1975]

yere thanks